Tricky Torque Problem: Solving for Tension and Understanding Beam Forces

[Arooj]

Homework Statement

The writing was just my reasoning, which I don't think is right.
Assume the beam is massless.

Homework Equations

T = Fd sin θ
F = mg
net force must be zero.
net torque must be zero.

The Attempt at a Solution

d * 1/2F + 980 = d * (sin 37) T
1/2 F + 980 = (sin 37) T
I don't know what to do from here.
I can mathematically solve the problem to an extent, but conceptually I don't understand how to do the problem. Can someone explain to me how to do the problem from scratch and the reasoning for steps (like what forces are acting on the beam)?

 

[Doc Al]

d * 1/2F + 980 = d * (sin 37) T
1/2 F + 980 = (sin 37) T

What's F? I thought the beam was massless.

To solve this sort of problem, apply the conditions for equilibrium. The net torque = 0 is one of them. What are the others?

 

[Arooj]

ah, so there is no F, meaning it would just be 980 = sin 37 T, where T is the tension.
net force = zero, forces left = forces right, forces up = forces down, clockwise torques = counterclockwise torques, x and y components of force may separately be set to 0. And T cos 37 would give the force exerted by the wall, I assume?

 

[Doc Al]

ah, so there is no F, meaning it would just be 980 = sin 37 T, where T is the tension.

Right.

net force = zero, forces left = forces right, forces up = forces down, clockwise torques = counterclockwise torques, x and y components of force may separately be set to 0. And T cos 37 would give the force exerted by the wall, I assume?

Exactly.

 

[Arooj]

And compression would thus be opposite to the force exerted by the wall, for a net force of zero?

 

[Doc Al]

And compression would thus be opposite to the force exerted by the wall, for a net force of zero?

Yes, the net force is zero. The force exerted by the wall is the compression.